The rang of an inverse function is the domain of the normal function
So you just gotta find the domain of that which is x is greater than or equal to 0
Answer:
t=6
Step-by-step explanation:
ground height = 0
(are you sure your formula is correct? isn't it - 16t²?)
if h=16t² +64t+192 is true then
16t² +64t+192 = 0
t² + 4t + 12 = 0
t = (-4 ± √(4² - 4*12)) / 2*1 = (-4 ± √-32) / 2 = -2 ± 2√-2
There is no solution of t
if it is h= - 16t² +64t+192
0 =- 16t² + 64t + 192
t² - 4t - 12 = 0
(t + 2) (t -6) = 0
t should be positive
t = 6 sec
Answer:

Step-by-step explanation:
(4/3)P -(4/3)A + A = B . . . . . . add A
(4P -A)/3 = B . . . . . . . . . . . . . simplify
Then the coordinates of point B are ...
B = (4(1, 6) -(-5, 3))/3 = (9, 21)/3
B = (3, 7)